What is the foci-(x^2}{16}+\frac{y^2)/9 =1 ?

The foci-(x^2}{16}+\frac{y^2)/9 =1 is (0,5),(0,-5)

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foci-(x^2}{16}+\frac{y^2)/9 =1

foci

- \frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1

(0, 5), (0, - 5)

Solution steps

(h, k + c), (h, k - c)

Calculate Hyperbola properties

- \frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 3, b = 4

(0, 0 + c), (0, 0 - c)

Compute

c :

c = 3^{2} + 4^{2} : 5

(0, 0 + 5), (0, 0 - 5)

Refine

(0, 5), (0, - 5)

4 x^{2} - 9 y^{2} = 144

\frac{( y - \frac{9}{2} ) ^{2}}{49} - \frac{( x - \frac{7}{2} ) ^{2}}{49} = 1

9 x^{2} - 4 y^{2} + 18 x - 27 = 0

\frac{x ^{2}}{8} - \frac{y ^{2}}{8} = 1

\frac{y ^{2}}{64} - \frac{x ^{2}}{49} = 1

What is the foci-(x^2}{16}+\frac{y^2)/9 =1 ?
The foci-(x^2}{16}+\frac{y^2)/9 =1 is (0,5),(0,-5)